cos x ( ln ( cos x ) ) 2 d y + ( sin x − 3 y ( sin x ) ln ( cos x ) ) d x = 0 \cos x(\ln (\cos x))^{2} d y+(\sin x-3 y(\sin x) \ln (\cos x)) d x= 0 cos x ( ln ( cos x ) ) 2 d y + ( sin x − 3 y ( sin x ) ln ( cos x )) d x = 0 cos x ( ln ( cos x ) ) 2 d y d x − 3 sin x ⋅ ln ( cos x ) y = − sin x \cos x(\ln (\cos x))^{2} \frac{d y}{d x}-3 \sin x \cdot \ln (\cos x) y=-\sin x cos x ( ln ( cos x ) ) 2 d x d y − 3 sin x ⋅ ln ( cos x ) y = − sin x d y d x − 3 tan x ln ( cos x ) y = − tan x ( ln ( cos x ) ) 2 \frac{d y}{d x}-\frac{3 \tan x}{\ln (\cos x)} y=\frac{-\tan x}{(\ln (\cos x))^{2}} d x d y − ln ( cos x ) 3 tan x y = ( ln ( cos x ) ) 2 − tan x d y d x + 3 tan x ln ( sec x ) y = − tan x ( ln ( sec x ) ) 2 \frac{d y}{d x}+\frac{3 \tan x}{\ln (\sec x)} y=\frac{-\tan x}{(\ln (\sec x))^{2}} d x d y + ln ( sec x ) 3 tan x y = ( ln ( sec x ) ) 2 − tan x I.F. = e ∫ 3 tan x ln ( sec x ) d x = ( ln ( sec x ) ) 3 \text { I.F. }=\mathrm{e}^{\int \frac{3 \tan x}{\ln (\sec x)} \mathrm{dx}}=(\ln (\sec x))^{3} I.F. = e ∫ l n ( s e c x ) 3 t a n x dx = ( ln ( sec x ) ) 3 y × ( ln ( sec x ) ) 3 = − ∫ tan x ( ln ( sec x ) ) 2 ( ln ( sec x ) ) 3 d x + C y \times(\ln (\sec x))^{3}=-\int \frac{\tan x}{(\ln (\sec x))^{2}}(\ln (\sec x))^{3} d x+C y × ( ln ( sec x ) ) 3 = − ∫ ( ln ( sec x ) ) 2 tan x ( ln ( sec x ) ) 3 d x + C y × ( ln ( sec x ) ) 3 = − 1 2 ( ln ( sec x ) ) 2 + C \mathrm{y} \times(\ln (\sec \mathrm{x}))^{3}=-\frac{1}{2}(\ln (\sec \mathrm{x}))^{2}+C y × ( ln ( sec x ) ) 3 = − 2 1 ( ln ( sec x ) ) 2 + C Given : x = π 4 , y = − 1 ln 2 \mathrm{x}=\frac{\pi}{4}, \mathrm{y}=-\frac{1}{\ln 2} x = 4 π , y = − l n 2 1
− 1 ln 2 × ( ln 2 ) 3 = − 1 2 × ( ln 2 ) 2 + C \frac{-1}{\ln 2} \times(\ln \sqrt{2})^{3}=-\frac{1}{2} \times(\ln \sqrt{2})^{2}+\mathrm{C} l n 2 − 1 × ( ln 2 ) 3 = − 2 1 × ( ln 2 ) 2 + C
⇒ − 1 8 ln 2 × ( ln 2 ) 3 = − 1 2 × 1 4 ( ln 2 ) 2 + C − 1 8 ( ln 2 ) 2 = − 1 8 ( ln 2 ) 2 + C ⇒ C = 0 \begin{aligned} & \Rightarrow \frac{-1}{8 \ln 2} \times(\ln 2)^{3}=\frac{-1}{2} \times \frac{1}{4}(\ln 2)^{2}+\mathrm{C} \\ & -\frac{1}{8}(\ln 2)^{2}=\frac{-1}{8}(\ln 2)^{2}+\mathrm{C} \\ & \Rightarrow \mathrm{C}=0 \end{aligned} ⇒ 8 ln 2 − 1 × ( ln 2 ) 3 = 2 − 1 × 4 1 ( ln 2 ) 2 + C − 8 1 ( ln 2 ) 2 = 8 − 1 ( ln 2 ) 2 + C ⇒ C = 0 ∴ y ( ln ( sec x ) ) 3 = − 1 2 ( ln ( sec x ) ) 2 + 0 \therefore y(\ln (\sec x))^{3}=\frac{-1}{2}(\ln (\sec x))^{2}+0 ∴ y ( ln ( sec x ) ) 3 = 2 − 1 ( ln ( sec x ) ) 2 + 0 y = − 1 2 ln ( sec x ) y=\frac{-1}{2 \ln (\sec x)} y = 2 ln ( sec x ) − 1 y = 1 2 ln ( cos x ) y=\frac{1}{2 \ln (\cos x)} y = 2 ln ( cos x ) 1 ∴ y ( π 6 ) = 1 2 ln ( cos π 6 ) \therefore y\left(\frac{\pi}{6}\right)=\frac{1}{2 \ln \left(\cos \frac{\pi}{6}\right)} ∴ y ( 6 π ) = 2 ln ( cos 6 π ) 1 = 1 2 ln ( 3 2 ) =\frac{1}{2 \ln \left(\frac{\sqrt{3}}{2}\right)} = 2 ln ( 2 3 ) 1 = 1 2 ( 1 2 ln 3 − ln 2 ) =\frac{1}{2\left(\frac{1}{2} \ln 3-\ln 2\right)} = 2 ( 2 1 ln 3 − ln 2 ) 1 = 1 ln 3 − ln 4 =\frac{1}{\ln 3-\ln 4} = ln 3 − ln 4 1